Let the first integer be x. Since the consecutive integers differ by 1, the other consecutive integers would be (x + 1), (x + 2), and (x + 3). So we have
x + x + 1 + x + 2 + x + 3 = 90
4x + 6 = 90
4x + 6 - 6 = 90 - 6
4x = 84
4x/4 = 84/4
x = 21
Thus, the third integer is 23 (21 + 2 = 23).
The let statement is: let the smallest of the three integers be x.
-1, 1 and 3
Suppose the middle integer is a. Then the three consecutive integers are a-1, a and a+1. Now, 3*a = (a-1)+(a+1) + 1 So 3a = 2a + 1 => a=1 So the three numbers are 0, 1 and 2
9, 11, and 139 + 2(11) + 3(13) = 9 + 22 + 39 = 70
Middle integer must be one-third of the sum, in this case 70. Other two are therefore 68 and 72
If the first of these consecutive integers is x, the second integer would be x + 1, and the third integer would be x + 2.Since the sum of the second and the third integer is 17, we can writex + 1 + x + 2 = 172x + 3 = 172x + 3 - 3 = 17 - 32x = 142x/2 = 14/2x = 7Thus, the consecutive integers are 7, 8, and 9.
10-11-12
Middle integer must be a third of 93 so integers are 30, 31 and 32.
The let statement is: let the smallest of the three integers be x.
Let the first consecutive integer be x. So that:the second integer is x + 1,the third integer is x + 2, andthe fourth integer is x + 3.We have:(x + 1) + (x + 3) = 1322x + 4 = 1322x = 128x = 64 the first integerThus, the four consecutive integers are 64, 65, 66, and 67.
-1, 1 and 3
The second integer is 64 (128/2), so the first is 62 and the third is 66.OrLet the 1st integer be x and the 3rd integer be y.Since even integers differ by 2, then we have:y = x + 4x + y = 128x + x + 4 = 1282x = 124x = 62So the integers are 62, 64, and 66, and their sum is 192
Suppose the middle integer is a. Then the three consecutive integers are a-1, a and a+1. Now, 3*a = (a-1)+(a+1) + 1 So 3a = 2a + 1 => a=1 So the three numbers are 0, 1 and 2
Let x=1st integer x+1=2nd integer x+2=3rd integer 5(x+x+1+x+2)=150 5(3x+3)=150 15x+15=150 15x=135 x=9 So you put what x is 1st integer (x)= 9 Since it's three consecutive integers, you add one to the 2nd integer 2nd integer (x+1)= 10 Then you do the same thing for the third integer 3rd integer (x+2)= 11
Consecutive integers could be thought of as counting numbers in a row. One of them is "lowest" and the next one will be one more than that, and the last one will be one more than the second one. The numbers 7, 8 and 9 and 46, 47 and 48 are each said to be three consecutive integers. You often come across a question that tells you that 3 consecutive integers add up to a value. Example 3 consecutive integers add to 6. What are the integers? Let the first integer be x the second is then x +1 the third x+2 add them x+x+1+x+2 =3x+3 and this would be equal to 6. we then have the equation 3x+3=6 3x=3 x=1 so the first integer is 1, the second would be 2 and the third 3
9, 11, and 139 + 2(11) + 3(13) = 9 + 22 + 39 = 70
Middle integer must be one-third of the sum, in this case 70. Other two are therefore 68 and 72